Invertible matrix / determinant

In summary, you proved that if B is invertible, then there exists a scalar cεC s.t. A + cB is not invertible.
  • #1
Wildcat
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Homework Statement


Let A,B ε Mnxn(C)
Prove that if B is invertible, then there exists a scalar cεC s.t. A + cB is not invertible. HINT:
examine det (A+cB)



Homework Equations





The Attempt at a Solution


I know that the det(A+cB)=0 since a non invertible matrix has det=0
I know B is invertible so I multiply the right side by B^-1 which gives det(AB^-1 + cIn)=0 where In is the nxn identity matrix. There is a theorem that says that a scalar is an eigenvalue of A if and only if det(A-cIn)=0.
can I choose c to be -c? does that show that A +cB is not invertible? Help!
 
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  • #2
So you need to find a c such that [tex]\det(AB^{-1}+cI_n )=0[/tex]. So this amounts to finding an eigenvalue c' of [tex]AB^{-1} [/tex] (or showing that such one exists) and setting c=-c'...
 
  • #3
You are on the right track. I think you are using the same letters to represent too many things. Do you care about the eigenvalues of A or of another matrix? Choosing c to be -c would imply that c=0, perhaps you need to use another symbol to represent the quantity that you'd use to choose your original c.
 
  • #4
can I rename AB^-1, D and say det(D-cIn)=0 I can find c for a specific 2x2 matrix, but I'm not sure how to do that in general
 
  • #5
Wildcat said:
can I rename AB^-1, D and say det(D-cIn)=0 I can find c for a specific 2x2 matrix, but I'm not sure how to do that in general

You're using c for too many things, it's better to write det(D-dIn)=0. The nature of the problem only requires that you know that at least one d exists, not that you have to find its value. Does every matrix have at least one eigenvalue?
 
  • #6
Yes? and an nxn would have at most n distinct eigenvalues right?
 
  • #7
Wildcat said:
Yes? and an nxn would have at most n distinct eigenvalues right?

Right, so you had the right idea before, you just want to express c in terms of one of the eigenvalues d so that the notation isn't confusing.
 
  • #8
so, if I use my original proof and add Let AB^-1 = D and c=-d then det(D-dIn)=0 would be enough? I don't have to show anything else?
 
  • #9
Wildcat said:
so, if I use my original proof and add Let AB^-1 = D and c=-d then det(D-dIn)=0 would be enough? I don't have to show anything else?

You can always follow your logic backwards to check that if c = -d, det(A+cB)=0. But your arguments looked fine to me.
 
  • #10
OK, Thank You!
 

FAQ: Invertible matrix / determinant

1. What is an invertible matrix?

An invertible matrix is a square matrix that has a unique solution for every system of linear equations. This means that the matrix can be "inverted" or reversed to solve for the original variables.

2. How do you determine if a matrix is invertible?

A matrix is invertible if its determinant is non-zero. The determinant is a numerical value that can be calculated using a specific formula. If the determinant is equal to zero, then the matrix is not invertible.

3. What is the significance of the determinant in an invertible matrix?

The determinant is a measure of how much the matrix "stretches" or "shrinks" space. In an invertible matrix, the determinant must be non-zero in order for the matrix to have a unique solution. A determinant of zero means that the matrix does not have a unique solution.

4. Can any matrix be inverted?

No, not all matrices are invertible. In order for a matrix to be invertible, it must be square (having the same number of rows and columns) and have a non-zero determinant. If these conditions are met, then the matrix can be inverted.

5. What is the inverse of an invertible matrix?

The inverse of an invertible matrix is another matrix that, when multiplied by the original matrix, results in the identity matrix (a square matrix with 1s along the diagonal and 0s everywhere else). This means that the inverse essentially "undoes" the original matrix's operations.

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